Help with linear algebra

Question

anonymous · Answer

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anonymous · Answer

If $\lambda_1=3+2i$, then you know $\lambda_2=3-2i$, since conjugate roots to the characteristic equation come in pairs. Not quite sure about $\lambda_3$ and $\lambda_4$ just yet.

phi · Answer

singular means   one lambda is 0

the trace of the matrix is the sum of the lambdas
so you can find the all 4

phi · Answer

btw  @loser,   isn't it   det ( A - lambda I ) = 0  for the characteristic equation ?

anonymous · Answer

@Loser66, that's not necessarily the char. equation. We don't know the other elements of the matrix to say that.

phi · Answer

you have
λ1=3+2i, λ2=3−2i, λ3= 0,  and λ1+λ2+λ3+λ4= 4

anonymous · Answer

Have do you know λ2=3−2i?

phi · Answer

your characteristic equation has real coefficients, so its complex roots come in complex conjugate pairs https://en.wikipedia.org/wiki/Complex_conjugate_root_theorem

anonymous · Answer

So the augment is:
1) If λ1=3+2i, then you know λ2=3−2i, since conjugate roots to the characteristic equation come in pairs
2) Because it is singular, one lambda is 0
3) The trace of the matrix is the sum of the lambdas
so I can find the all 4

phi · Answer

yes,  and you know the trace is 4  (all 1's on the diagonal)

anonymous · Answer

Thank you @phi 

I have another problem i cant solve. Will you give it a try?